higher order determinants
TRANSCRIPT
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2.2. Higher-Order Determina nts
The 1
1 matrix [a] is invertible exactly when a
= 0.
The 2 2 matrix
a bc d
is invertible exactly when
ad bc = 0. What about a 3 3 matrix? Is there someshort of expression which will determine when such a
matrix is invertible?The answer is yes (and the answer for larger squarematrices is also yes), and it is called the d e t e r m i n a n t of the matrix, but the formula gets very complicated
very fast as the matrix gets bigger; it is easier towork on an individual basis.
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Lets proceed one step at a time, starting with them i n o r s of a matrix A.
The determinant of the matrix obtained by removingall the entries in the ith row and the jth column of Ais called the (i, j) t h m i n o r o f A and is denoted Mi,j.
This assumes you know how to find the determinant
of a smaller matrix!
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If A =
2 4 34 8 33
6 1
, then M1,2, (1, 2)th minor of A, is
the determinant of
2 4 34 8 33
6 1
=
4 33 1
, which is 4 1 3 (3) = 13.
M2,1 is the determinant of
2 4 34 8 33
6 1
=
4 3
6 1
, which is (4) 1 (6) 3 = 14.
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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which
changes the signs of some of the minors:
+ +
+
+ +
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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which
changes the signs of some of the minors:Minors
+ +
+
+ + 26 13 0
14
7 0
36 18 0
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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which
changes the signs of some of the minors:Minors Cofactors
+ +
+
+ + 26 13 0
14
7 0
36 18 0 26 13 0
14
7 0
36 18 0
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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which
changes the signs of some of the minors:Minors Cofactors
+
+
+
+ +
26 13 014
7 0+ 36 18 0
26 13 0
14
7 0+ 36 18 0
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The (i, j) c o f a c t o r o f A, denoted Ci,j, is (1)i+jMi,j.The (1)i+j creates a checkerboard pattern, which
changes the signs of some of the minors:Minors Cofactors
+
+
+
+ +
26+
13 0+
14
7 0
36 18 0
26
13 0
14
7 0
36 + 18 0
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Now (at last) we can find the determinant of A,which is defined to be
A1,1C1,1 + A1,2C1,2 + A1,3C1,3 + + A1,nC1,n.
The determinant of a matrix A is denoted by |A| ordet A.
Here, the determinant is (2)(26)+(4)(13) + (3)(0) =0 .
This method is called E x p a n s i o n b y M i n o r s .
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A more interesting example, with some of the detailsomitted, is:
2 7 31 3 23 7 9
= 2 +
3 27 9
+ 7
1 23 9
+ 3 +
1 33 7
= 2 (3 9 2 7) 7 (1 9 2 3)+ 3 (1 7 3 3)
= 1
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As you can see, calculating the determinant of a 3 3matrix requires calculating the determinant of 3 2 2
matrices.
Calculating the determinant of a 4 4 matrix requirescalculating the determinant of 4 3 3 matrices, eachof which requires the determinants of 3 2
2 matrices.
This makes a total of 12 2 2 matrices.Calculating the determinant of a 5 5 matrix requirescalculating the determinant of 5 4 4 matrices, which
will require the determinant of 60 2 2 matrices.
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In general, the determinant of an n n matrixrequires the determinants of
1
2n! 2 2 matrices. This
is not an efficient procedure! (n! = 1 2 . . . n. 10! isaround 3 million, 70! is bigger than a googol.)
So how can we cut down the computation time?
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First of all, we can expand along any row and getthe same answer, not just the first one. Also, we canexpand along any column. So if some row or columnhas a lot of 0s in it, we can cut down the number ofcomputations.
We need to obey the checkerboard pattern, sothe first determinant might be subtracted insteadof being added.
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
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3 2 8 04 17 1
2
0 0 40
6 1 7 0
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 0
4 17 10 0 4
6 1 7
+ 2 +
3 2 80 0 4
6 1 7
+ 0
3 2 84 17 16 1
7
+ 0 +
3 2 84 17 10 0 4
Remember:
+ +
+ ++ +
+
+
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 2
3 2 80 0 4
6 1 7
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 2
3 2 80 0 4
6 1 7
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 2
3 2 80 0 4
6 1 7
= 2
0
2 81 7
+ 0 +
3 86 7
+ 4
3 26 1
Remember: + + +
+ +
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 2
3 2 80 0 4
6 1 7
= 2 4
3 26 1
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3 2 8 04 17 1 20 0 4 0
6 1 7 0
= 2
3 2 80 0 4
6 1 7
= 2 4
3 26 1
= 2 4 (3 1 2 6) = 72
Only one 2 2 determinant had to be calculated here!
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Now suppose we have an upper triangular matrix*Maybe we want to find the determinant of:
1 3 2 4 0 2 1 2 30 0 3
2 2
0 0 0 5 10 0 0 0 8
* An u p p e r t r i a n g u l a r m a t r i x is a matrix A whereAi,j = 0 whenever j < i.
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1 3 2 4 0 2 1 2 30 0 3
2 2
0 0 0 5 10 0 0 0 8
= 1
2 1 2 30 3
2 2
0 0 5 10 0 0 8
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1 3 2 4 0 2 1 2 30 0 3
2 2
0 0 0 5 10 0 0 0 8
= 1
2
3
2 2
0 5 1
0 0 8
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1 3 2 4 0 2 1 2 30 0 3
2 2
0 0 0 5 10 0 0 0 8
= 1
2
3
5 1
0 8
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1 3 2 4 0 2 1 2 30 0 3
2 2
0 0 0 5 10 0 0 0 8
= 1
2
3
5
|8
|
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1 3 2 4 0
21 2 3
0 03
2 2
0 0 0 5 10 0 0 0 8
= 1
2
3
5
8
T h e d e t e r m i n a n t o f a n u p p e r ( o r l o w e r ) t r i a n g u l a r m a t r i x
i s t h e p r o d u c t o f t h e e n t r i e s o n t h e d i a g o n a l .
We could use this fact, if we knew how to get amatrix into an upper triangular form . . .